Let the solution curve of the differential equation $xdy−ydx=\sqrt{{x}^{2}+{y}^{2}}dx,x>0$, $y(1)=0,$ be $y=y(x)$. Then $y(3)$ is equal to

[JEE Main 2026, 22 Jan (Shift 1)]

Question

Let the solution curve of the differential equation $xdy−ydx=\sqrt{{x}^{2}+{y}^{2}}dx,x>0$, $y(1)=0,$ be $y=y(x)$. Then $y(3)$ is equal to [JEE Main 2026, 22 Jan (Shift 1)]

YESPYQ · Accepted Answer

Options are free to see. Unlock the correct answer and full explanation with PYQ Pass.

Let the solution curve of the differential equation \(xdy−ydx=\sqrt{{x}^{2}+{y}^{2}}dx,x>0\), \(y(1)=0,\) be \(y=y(x)\).…

Practice more Board Maths PYQs